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Math Help - Question involving a collection of closed intervals

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    Question involving a collection of closed intervals

    Let F be the collection of closed intervals A_n=[1/n, 1-1/n] for n=3,4,5,.... What do you notice about \bigcup F? Is it closed, open, both, or neither?

    I'm afraid the wording and available material has got me stumped on this, and being sick at the moment doesn't help.
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    Notice that \left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1.
    Can either 1 or 0 be included?
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    Quote Originally Posted by Plato View Post
    Notice that \left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1.
    Can either 1 or 0 be included?
    I should note, the original question said that n=1,2,3,..., but our Prof. told us to change it to n=3,4,5,... when he realized that when n=1, A_n=[1,0] and when n=2, A_n=[1/2,1/2] (or at least that's what I heard him say). I wrote the question word-for-word (besides the change our Prof. gave us), so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
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    Quote Originally Posted by Runty View Post
    so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
    That is the whole point: neither 0 nor 1 can be included.
    So what is the union?
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    Quote Originally Posted by Plato View Post
    That is the whole point: neither 0 nor 1 can be included.
    So what is the union?
    I am presently guessing, but here is my answer thus far.

    \bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]

    My guess so far indicates that the union is both open and closed, though I could easily be wrong.
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    Quote Originally Posted by Runty View Post
    I am presently guessing, but here is my answer thus far.

    \bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]
    For any k does 1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?
    If not how can the union be (0,1]?
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    Quote Originally Posted by Plato View Post
    For any k does 1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?
    If not how can the union be (0,1]?
    Point taken. I guess it's a completely open interval, then.

    EDIT: One more thing, I've looked at some work done on this in another forum, for which I will provide the link.

    http://www.mymathforum.com/viewtopic...=15805&start=0

    Is the work found here accurate?
    Last edited by Runty; September 19th 2010 at 12:31 PM.
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